# Build Lowest Number by Removing n digits from a given number

Given a string ‘str’ of digits and an integer ‘n’, build the lowest possible number by removing ‘n’ digits from the string and not changing the order of input digits.

Examples:

```Input: str = "4325043", n = 3
Output: "2043"

Input: str = "765028321", n = 5
Output: "0221"

Input: str = "121198", n = 2
Output: "1118" ```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is based on the fact that a character among first (n+1) characters must be there in resultant number. So we pick the smallest of first (n+1) digits and put it in result, and recur for remaining characters. Below is complete algorithm.

```Initialize result as empty string
res = ""
buildLowestNumber(str, n, res)
1) If n == 0, then there is nothing to remove.
Append the whole 'str' to 'res' and return

2) Let 'len' be length of 'str'. If 'len' is smaller or equal
to n, then everything can be removed
Append nothing to 'res' and return

3) Find the smallest character among first (n+1) characters
of 'str'.  Let the index of smallest character be minIndex.
Append 'str[minIndex]' to 'res' and recur for substring after
minIndex and for n = n-minIndex

buildLowestNumber(str[minIndex+1..len-1], n-minIndex).```

Below is C++ implementation of above algorithm.

```// C++ program to build the smallest number by removing n digits from
// a given number
#include<iostream>
using namespace std;

// A recursive function that removes 'n' characters from 'str'
// to store the smallest possible number in 'res'
void buildLowestNumberRec(string str, int n, string &res)
{
// If there are 0 characters to remove from str,
// append everything to result
if (n == 0)
{
res.append(str);
return;
}

int len = str.length();

// If there are more characters to remove than string
// length, then append nothing to result
if (len <= n)
return;

// Find the smallest character among first (n+1) characters
// of str.
int minIndex = 0;
for (int i = 1; i<=n ; i++)
if (str[i] < str[minIndex])
minIndex = i;

// Append the smallest character to result
res.push_back(str[minIndex]);

// substring starting from minIndex+1 to str.length() - 1.
string new_str = str.substr(minIndex+1, len-minIndex);

// Recur for the above substring and n equals to n-minIndex
buildLowestNumberRec(new_str, n-minIndex, res);
}

// A wrapper over buildLowestNumberRec()
string buildLowestNumber(string str, int n)
{
string res = "";

// Note that result is passed by reference
buildLowestNumberRec(str, n, res);

return res;
}

// Driver program to test above function
int main()
{
string str = "121198";
int n = 2;
cout << buildLowestNumber(str, n);
return 0;
}
```

Output:

`1118`

Below is an optimised code in C++ contributed by Gaurav Mamgain

```// C++ program to build the smallest number by removing
// n digits from a given number
#include<bits/stdc++.h>
using namespace std;

void insertInNonDecOrder(deque<char> &dq, char ch)
{
// if container is empty , insert the current digit
if (dq.empty())
dq.push_back(ch);

else
{
char temp = dq.back();

// Keep removing digits larger than current digit
// from the back side of deque
while( temp > ch && !dq.empty())
{
dq.pop_back();
if (!dq.empty())
temp = dq.back();
}
dq.push_back(ch); // insert the current digit
}
return;
}

string buildLowestNumber(string str, int n)
{
int len = str.length();

// deleting n digits means we need to print k digits
int k = len - n;

deque<char> dq;
string res = "";

// leaving rightmost k-1 digits we need to choose
// minimum digit from rest of the string and print it
int i;
for (i=0; i<=len-k; i++)

// insert new digit from the back side in
// appropriate position and/ keep removing
// digits larger than current digit
insertInNonDecOrder(dq, str[i]);

// Now the minimum digit is at front of deque
while (i < len)
{
// keep the minimum digit in output string
res += dq.front();

// remove minimum digit
dq.pop_front();

// Again insert new digit from the back
// side in appropriate position and keep
// removing digits larger than current digit
insertInNonDecOrder(dq, str[i]);
i++;
}

// Now only one element will be there in the deque
res += dq.front();
dq.pop_front();
return res;
}

// Driver program to test above function
int main()
{
string str = "765028321";
int n = 5;
cout << buildLowestNumber(str, n)<< endl;
return 0;
}
// This code is contributed by Gaurav Mamgain
```

Output:

`0221`

Time Complexity: O(n)
Space Complexity: O(n)